dictionary - Merge a dict in Python using 1 dict as base -


this question has answer here:

i looking feedback on python code. trying merge 2 dictionaries. 1 of dictionaries controls structure , default values, second dictionary overwrite default values when applicable.

please note looking following behaviour:

  • keys present in other dict should not added
  • nested dicts should taken account

i wrote simple function:

def merge_dicts(base_dict, other_dict):     """ merge 2 dicts      ensure base_dict remains , overwrite info other_dict     """     out_dict = dict()     key, value in base_dict.items():         if key not in other_dict:             # use base             nvalue = value         elif isinstance(other_dict[key], type(value)):             if isinstance(value, type({})):                 # new dict myst recursively merged                 nvalue = merge_dicts(value, other_dict[key])             else:                 # use others' value                 nvalue = other_dict[key]         else:             # error due difference of type             raise typeerror('the type of key {} should {} (currently {})'.format(                 key,                 type(value),                 type(other_dict[key]))             )         out_dict[key] = nvalue     return out_dict

i sure can done more beautifully/pythonic.

"pythonicness" hard measure assess, here take on it:

def merge_dicts(base_dict, other_dict):     """ merge 2 dicts      ensure base_dict remains , overwrite info other_dict     """     if other_dict none:         return base_dict     t = type(base_dict)     if type(other_dict) != t:         raise typeerror("mismatching types: {} , {}."                         .format(t, type(other_dict)))     if not issubclass(t, dict):         return other_dict     return {k: merge_dicts(v, other_dict.get(k)) k, v in base_dict.items()}

example:

merge_dicts({"a":2, "b":{"b1": 5, "b2": 7}}, {"b": {"b1": 9}}) >>> {'a': 2, 'b': {'b1': 9, 'b2': 7}}

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